Question

Two bodies undergo one-dimensional, perfectly inelastic, head-on collision. Derive an expression for the loss in the kinetic energy.

Answer 1

i. Let two bodies A and B of masses m1 and m2 move with initial velocity \(\overrightarrow{u}_1\) and \(\overrightarrow{u}_2\), respectively such that particle A collides head on with particle B i.e., u₁ > u₂.

ii. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\overrightarrow{\text{v}}\) after the collision along the same straight line.

loss in kinetic energy = total initial kinetic energy – total final kinetic energy.

iii. By the law of conservation of momentum,

m₁u₁ + m₂u₂ = (m₁ + m₂)v

∴ v = \(\frac{m_1u_1+m_2u_2}{m_1+m_2}\)

iv. Loss of Kinetic energy,

v. Both the masses and the term (u₁ – u₂)² are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as e = 0, the loss is maximum.

Answer 2

USE

1) IF ELASTIC COLLISION THEN ENERGY CONSERVATION

K.E.(INITIAL)=K.E.(FINAL)

2) IN INELASTIC USE MOMENTUM CONSERVATION